Hihocoder 1273 清理海报(判断矩形相交)

题意:

中文题意

分析:

$如果i可以被j覆盖就从i向j连一条边,判断矩形相交看代码吧$
$覆盖的情况可以状压矩形的四个角,判断能不能撕开$
$时间复杂度O(n^2)$

代码:

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//
// Created by TaoSama on 2016-03-13
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e3 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
#define y1 asdasdasd
int w, h, n;
int x1[N], y1[N], x2[N], y2[N];
bool vis[N];
vector<int> G[N];
bool intersect(int i, int j) {
int minx = max(x1[i], x1[j]);
int maxx = min(x2[i], x2[j]);
int miny = max(y1[i], y1[j]);
int maxy = min(y2[i], y2[j]);
if(minx >= maxx || miny >= maxy) return false;
return true;
}
bool inside(int x, int y, int i) {
return x > x1[i] && x < x2[i] && y > y1[i] && y < y2[i];
}
//i被j包含
int cover(int i, int j) {
int ret = 0;
if(inside(x1[i], y1[i], j)) ret |= 1;
if(inside(x1[i], y2[i], j)) ret |= 2;
if(inside(x2[i], y2[i], j)) ret |= 4;
if(inside(x2[i], y1[i], j)) ret |= 8;
return ret;
}
void dfs(int st, int u, int& sta) {
vis[u] = true;
for(int v : G[u]) {
if(vis[v]) continue;
sta |= cover(st, v);
dfs(st, v, sta);
}
}
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
scanf("%d%d%d", &w, &h, &n);
for(int i = 1; i <= n; ++i) {
scanf("%d%d%d%d", x1 + i, y1 + i, x2 + i, y2 + i);
}
for(int i = 1; i <= n; ++i) {
for(int j = i + 1; j <= n; ++j)
if(intersect(i, j)) G[i].push_back(j);
}
// for(int i = 1; i <= n; ++i){
// printf("%d: ", i);
// for(int v : G[i]) printf("%d ", v); puts("");
// }
pair<int, int> ans = make_pair(0, 0);
for(int i = 1; i <= n; ++i) {
memset(vis, false, sizeof vis);
int sta = 0;
dfs(i, i, sta);
if(sta == 15) continue;
int cnt = 0;
for(int j = 1; j <= n; ++j) cnt += vis[j];
if(cnt > ans.first) ans = make_pair(cnt, i);
}
printf("%d %d\n", ans.first, ans.second);
return 0;
}


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